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STRUCTURE OF Mo-92, Mo-94, Mo-95, Mo-96, Mo-97, Mo-98, AND Mo-100
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. .STRUCTURE OF Mo-92, Mo-94, Mo-96, Mo-98 AND Mo-100 WITH S=0 In general all the above nuclides of Mo with 42 deuterons have a structure of high symmetry. Using the diagram of Zr with 40 deuterons of opposite spins one sees that the additional deuterons like p41n41 of S =-1 and p42n42 of S=+1 give the structure of high symmetry when they fill the symmetrical blank positions existing in front of n39p39 and in front of p38n38 respectively. For the stable structure they receive a number of extra neutrons as 8, 10, 12, 14, and 16 with opposite spins respectively. Under these conditions all the above nuclides give S=0 STRUCTURE OF Mo-95 AND Mo-97 WITH S = +5/2 In these stable structures the same 42 deuterons receive a number of extra neutrons as 11 and 13 respectively (0dd number which gives a spin different than zero). Under this condition the p41n41 changes the spin from S= -1 to S=+1 in order to fill the symmetrical position existing behind the p40n40.Since this change gives S=+2 one concludes that the first nuclide receives 6 extra neutrons of positive spins and five extra neutrons of negative spins. That is for the stable structure of Mo-95 one gets S = +2 + 6(+1/2) + 5(-1/2) = +5/2 Similarly for the stable structure of Mo-97 with 13 extra neutrons one concludes that the number of extra nucleons of positive spins is 7, while the number of extra neutrons of negative spins is 6. That is S = +2 + 7(+1/2) + 6(-1/2) = +5/2 ' ' DIAGRAM OF Zr WITH 40 DEUTERONS , 8 EXTRA n AND 12 EXTRA (n) ' '''Here the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Also the 12 extra neutrons (n) with weak bonds of six planes and the 4 extra neutrons n of strong bonds near the p37 and p38 are not shown. You can see only 4 extra neutrons n of strong bonds existing under the p21 and p22 and over the p31 and p32. ' ''' n40.........p40.......n ' n........p38..........n38 Horizontal squarewith n ' ' n31………p12..........n12........p32' ' p31....... n11.........p11…… n32 Sixth horizontal plane' ' p29....... n10.........p10…….... n30' ' n29………..p9..........n9 …….p30 Fifth horizontal plane' ' n27.........p8..........n8...........p28' ' p27.........n7..........p7........n28 Fourth horizontal plane' ' p25.........n6.........p6..........n26' ' n25……….p5........n5……….p26 Third horizontal plane' ' n23………p4........n4………….p24' ' p23……..n3………p3………..n24 Second horizontal plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First horizontal plane' ' n'.........p37 ......n37 ' ' n39......p39.........n Horizontal square with n ' ' ' '''TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ''' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' Category:Fundamental physics concepts